Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $z = \dfrac{k - 3}{-6k - 6} \div \dfrac{k^2 - 6k + 9}{k - 3} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $z = \dfrac{k - 3}{-6k - 6} \times \dfrac{k - 3}{k^2 - 6k + 9} $ First factor the quadratic. $z = \dfrac{k - 3}{-6k - 6} \times \dfrac{k - 3}{(k - 3)(k - 3)} $ Then factor out any other terms. $z = \dfrac{k - 3}{-6(k + 1)} \times \dfrac{k - 3}{(k - 3)(k - 3)} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac{ (k - 3) \times (k - 3) } { -6(k + 1) \times (k - 3)(k - 3) } $ $z = \dfrac{ (k - 3)(k - 3)}{ -6(k + 1)(k - 3)(k - 3)} $ Notice that $(k - 3)$ appears twice in both the numerator and denominator so we can cancel them. $z = \dfrac{ \cancel{(k - 3)}(k - 3)}{ -6(k + 1)\cancel{(k - 3)}(k - 3)} $ We are dividing by $k - 3$ , so $k - 3 \neq 0$ Therefore, $k \neq 3$ $z = \dfrac{ \cancel{(k - 3)}\cancel{(k - 3)}}{ -6(k + 1)\cancel{(k - 3)}\cancel{(k - 3)}} $ We are dividing by $k - 3$ , so $k - 3 \neq 0$ Therefore, $k \neq 3$ $z = \dfrac{1}{-6(k + 1)} $ $z = \dfrac{-1}{6(k + 1)} ; \space k \neq 3 $